23 BEPORT — 1872. 



On Tricliametral Quartan Curves. By F. W. ISTewman". 



Problem. To find the conditions that a Quartan may have 3 Diameters. 

 That it may have one, the equation must admit the form 



where X„ means a function of x of the nth degree. 

 Let P = a;- +y^j then we may write 



This form will not be changed if we change the origin to any point in the axis 

 of y ; hence, if there be a second diameter, we may may suppose it to pass through 

 the origin, which we treat as a Pole, making a; = rcosi^, y = r sin i|/-. 



Then rtH+(Ar'cos= \;^+Brcos x|r+C)r2 = 



/i?'* cos*\|r -if Ir^ cos^ -v//- + vir^ cos^ ^ + nr cos ■^+23 = 0, 



which by the routine of trigonometry is expressible as 



^a-^k+i(A-k) cos2ylr-i7c cos 4a|^|H 

 + { (B - f cos V^ - j; cos 3\|^ } r^ 

 + 1 (C — |?n) — |m cos 2\j/^r-=nr cos ^/'+p■ 



This is the equation of every Quartan which has so much as one diameter. 



In order that the line expressed by x/' = y may be a new diameter, it is necessary and it 

 suffices that the same equation should result by substituting \|/- = y+(B, and\jr=y—eo, 

 where y is a definite constant, r, a the variables of the equation. Put yj/ = y+to ; then 

 in order that + may give the same result, the terms concerned must vanish in the 

 coefficients of r*, r'^. r'^, r separatelJ^ It must be observed that the assumption 

 y = or y=180° is useless; and y = 90° leads us to two rectangular diameters, not 

 to three. Hence we must avoid to suppose sin y=0 or sin2y=0. 



Now (1) nr sin y . sin <b=0, .•. w=0 ; 



(2) m sin 2y sin 2a) = 0, .*. m=Q ; 



(3) in the coefficient of r^, we need at once 

 (B — |^)8in y sin cl>=0 ; \l sin 3y . sin So = 0. 



It is useless to suppose ^=0, B=0 ; for this, joined with m=Q, w=0, reduces the 

 equation to the Doubly Diametral. Hence our only useful results are 



sin3y=0, B=f/; which leave B and/ finite. 



(4) (A — Jc) sin 2y . sin 2ii=0,k sin 4y sin 4ci)=0. 



We cannot make sin 4y=0, since we already require sin3y = 0. Hence nothing 

 remains but k=0, A=0. 

 Thus the original equation is reduced to 



ar''+(Ba:+C)r==fBar'+p; Qi) 



and from sin3y=0we get i)6'o new diameters, defined by y=60° and y = 180°. 

 Thus the problem is solved. 



Originally, the assumption «=0 would have left our monodiametral curve still 

 a Quartan. But after supposing A = and /i-=0, we cannot make a also = without 

 reducing the equation to a Tertian. In fact it is easy to show that the conditions 

 here investigated yield the known Tertian Tnjiiyawhen-we add the assumption a = 0. 



Writing x = r cos yjr, 4a'^ = r' (cos 3\//' + 3 cos i/^), we find 



■ ar*+Cr2=iB/-»cos3>/r+i?, (t) 



which is the most general Polar Equation of Tridiametral Quartans. 

 Again, solving (/;) for r-, and making a=l, since a must be finite, 



Thus the general equation to rect. coords, has the form 



2/=+a:=+B'.r+C'==V{|BV+(B'a-+CT+E}, .... (;) 

 which has 3 Parameters, 



